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The specific rotation \(\left[(\alpha)^{25^{\circ}}\right.\) D \(]\) of \(L\) -alanine is \(+1.8\). Calculate the observed rotation of a \(1.10 \mathrm{M}\) solution in a 2.5-dm polarimeter tube at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The observed rotation \(\alpha\) of the L-alanine solution at \(25^{\circ}\,\mathrm{C}\) can be calculated using the formula \(\alpha = \left[(\alpha)^{25^{\circ}}\right]_\text{D} \cdot c \cdot l\), where \(\left[(\alpha)^{25^{\circ}}\right]_\text{D} = +1.8^\circ\), \(c = 1.10 \,\text{M}\), and \(l = 2.5\, \text{dm}\). Plugging in these values and performing the calculations yields an observed rotation of \(4.95^\circ\).

Step by step solution

01

Note down the given information

We are given the following information: - Specific rotation of L-alanine, \(\left[(\alpha)^{25^{\circ}}\right]_\text{D}\) = \(+1.8^\circ\) - Concentration of the solution, \(c = 1.10 \,\text{M}\) - Polarimeter tube length, \(l = 2.5\, \text{dm} = 25\, \text{cm}\) (1dm = 10cm) - Temperature, \(T = 25^{\circ}\,\text{C}\)
02

Use the formula for specific rotation

We can use the formula for specific rotation, which relates the observed rotation, specific rotation, concentration, and path length: \[ [\alpha]_\text{D} = \frac{\alpha}{c \cdot l} \] Where: - \([\alpha]_\text{D}\) is the specific rotation - \(\alpha\) is the observed rotation - \(c\) is the concentration of the solution - \(l\) is the path length in decimeters We need to find the observed rotation, \(\alpha\), so we can rearrange the formula to solve for \(\alpha\): \[ \alpha = [\alpha]_\text{D} \cdot c \cdot l \]
03

Plug in the given values and calculate the observed rotation

Now, we'll plug in the given values to find the observed rotation: \[ \alpha = (+1.8^\circ) \cdot (1.10\, \text{M}) \cdot (2.5\, \text{dm}) \] Calculating the result: \[ \alpha = 4.95\,^\circ \] So, the observed rotation of the L-alanine solution in a 2.5-dm polarimeter tube at \(25^{\circ}\,\text{C}\) is \(4.95^\circ\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Optical Activity
Optical activity is a fundamental property displayed by chiral molecules, which refers to their ability to rotate the plane of polarized light. When light passes through a substance containing chiral molecules, the orientation of the light wave can change direction, either to the right (dextrorotary) or to the left (levorotary), depending on the structure of the chiral molecules.

This behavior is significant in chemistry and pharmacology, as the direction and degree of rotation can affect both the reactions the molecules undergo and their biological activity. In practical terms, the specific rotation, noted as \( [\alpha]_D \), provides us with a standardized measure of a compound's optical activity and directly correlates with the molecular configuration of that substance.
Polarimetry
Polarimetry is the technique used to measure the degree of optical activity of a substance. A device called a polarimeter facilitates this measurement. The concept is simple: a beam of polarized light is passed through a sample, and the device measures the angle by which the plane of polarized light has been rotated upon emerging from the sample.

The polarimeter plays a crucial role in analyzing substances in chemical and pharmaceutical industries, helping to identify substances, determine their purity, and even estimate their concentration. The observed rotation provided by a polarimeter isn't just a raw number; it can be affected by several factors including the concentration of the optically active compound, the length of the polarimeter tube, the temperature at which the measurement is taken, and the wavelength of the light used.
Solution Concentration
Solution concentration is a measure of the amount of a solute present in a given quantity of solvent. It is an important factor in the context of polarimetry because the optical activity, as measured by specific rotation, depends on how concentrated the solution is.

The more concentrated the solution, the greater the amount of chiral molecules that the light will encounter as it passes through, which can result in a higher degree of rotation. This relation is quantified in the formula for specific rotation \( [\alpha]_D = \frac{\alpha}{c \cdot l} \), where \( c \), the concentration of the solution, is directly proportional to \( \alpha \), the observed rotation. This is clearly demonstrated in the step-by-step solution of L-alanine's specific rotation calculation, where knowing the concentration is crucial to determining the observed rotation.
Polarimeter Tube Length
The length of the polarimeter tube, denoted as \( l \), is a vital factor in the calculation of optical rotation. The path length dictates the amount of optically active medium that the polarized light travels through. A longer tube length means the light interacts with more of the substance, and therefore, typically results in a greater rotation.

In the case of our exercise, the polarimeter tube length is provided in decimeters (dm), which must be converted to centimeters (cm) when plugging into the formula, as it is the standard unit of length in these measurements. The length of the polarimeter tube directly multiplies the specific rotation and concentration in the formula to yield the observed rotation, illustrating its direct impact on the final measured optical activity.

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Most popular questions from this chapter

Paper electrophoresis at \(\mathrm{pH} 6.0\) was carried out on a mixture of glycine, alanine, glutamic acid, lysine, arginine and serine, (a) Which compound moved toward the anode? (b) Which moved toward the cathode? (c) Which remained at the origin?

A sample of polymer contains \(0.50\) mole fraction with molecular weight 100,000 and \(0.50\) mole fraction with molecular weight 200,000 . Calculate (a) the number average molecular weight, \(\mathrm{M}_{\mathrm{n}}\) and (b) the weight average molecular weight, \(\mathrm{M}_{\mathrm{w}}\).

At \(\mathrm{pH}\) values of 4 to 9 , natural amino acids exist as polar or zwitter ions: \(\mathrm{H}_{2} \mathrm{~N}^{+} \mathrm{CHZCO}^{-}{ }_{2}\). At a pH of 12 , what would be the predominant ionic type? What would it be at a \(\mathrm{pH}\) of 2? Can the uncharged molecular form, \(\mathrm{H}_{2} \mathrm{NCHZCO}_{2} \mathrm{H}\) ever predominate at any \(\mathrm{pH}\) ?

A chemist has an \(\mathrm{E}\). coli bacterium cell of cylindrical shape. It is \(2 \mu\) (microns) long and ly in diameter and weighs \(2 \times 10^{-12} \mathrm{~g}\). (a) How many lipid molecules are present, assuming their average molecular weight to be 700 and lipid content to be \(2 \%\) ? (b) If the cell contains 15,000 ribosomes, what percent of the volume do they occupy? You may assume them to be spherical with a diameter of \(180 \AA\); also \(10_{4} \AA=1\) micron.

In the Citric Acid Cycle (also known as the Krebs Cycle), citrate is converted to isocitrate. From the algebraic sign of \(\Delta \mathrm{G}^{\circ}\) for the isomerization, what is the favored or spontaneous direction of the reaction? Calculate \(\Delta \mathrm{G}^{\circ}\) for the following reaction and explain what bearing this reaction will have on the isomerization of citrate and the operation of the Krebs Cycle: Isocitrate \(^{3-}+(1 / 2) \mathrm{O}_{2}(\mathrm{~g})+\mathrm{H}^{\mathrm{f}} \rightleftarrows\) a ketoglutarate \(^{2-}+\mathrm{H}_{2} \mathrm{O}(\ell)+\) (g) \(\Delta \mathrm{G}_{\mathrm{f}}\) in \((\mathrm{Kcal} / \mathrm{mole})\) are for citrate \({ }^{3-}=-279.24\); \(\mathrm{CO}_{2}(\mathrm{~g}\) isocitrate \(^{3-}=-277.65 ; \mathrm{H}^{\mathrm{f}}=0, \mathrm{O}_{2}=0\) \(\alpha\) -ketoglutarate \(^{2-}=-190.62 ; \mathrm{H}_{2} \mathrm{O}(\ell)=-56.69\) \(\mathrm{CO}_{2}(\mathrm{~g})=-94.26\)

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